Question

A ring of the radius $r$ is suspended from a point on its circumference. If the ring is made to oscillate in the plane of the figure, then the angular frequency of these small oscillations is

• A. $\sqrt{\frac{\text{2}\text{g}}{\text{3}\text{r}}}$
• B. $\sqrt{\frac{\text{g}}{\text{2}\text{r}}}$
• C. $\sqrt{\frac{\text{5}\text{g}}{\text{4}\text{r}}}$
• D. $\sqrt{\frac{\text{4}\text{g}}{\text{3}\text{r}}}$

Answer 1

Answer: (B)

It is a physical pendulum, the time period of which is,
$T=2\pi \sqrt{\frac{I}{mgl}}$
Here, $I=$ moment of inertia of the ring about the point of suspension
$\begin{array}{l}=m{r}^2+m{r}^2\\ =2m{r}^2\end{array}$
and $I=$ distance of the point of suspension from the centre of gravity
$=r$
$\therefore T=2\pi \sqrt{\frac{2m{r}^2}{mgr}}=2\pi \sqrt{\frac{2r}{g}}$
$\therefore$ Angular frequency $\omega =\frac{2\pi }{T}$
or $\omega =\sqrt{\frac{g}{2r}}$